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{1,2,3,...}, and d

{1,2,3,...}, and d

In general there are are

In general there are are a total of 6! different ways to arrange the EE students. Since no CEstudents can stand together, the positions of the CE students must be chosen from amongthe seven “blanks” ( ) above. There are ( 74)ways to choose the positions, and then 4! waysto permute the students within the chosen positions. Thus there are a total of( 76! · · 4! =4)6!7! = 6048003!distinct arrangements.10. Let x i , i ∈ {1, 2, . . . , 10}, denote the number of muffins of type i.(a) We are interested in the number of different solutions to the equationx 1 + x 2 + · · · + x 10 = 6,subject to x i ≥ 0 for all i. This is equivalent to the number of different arrangements of6 objects and 9 dividers, of which there are( ) 15= 15!6 6!9! = 5005possibilities.(b) Suppose x 1 denotes the number of blueberry muffins. We are interested in the numberof different solutions to the equationx 1 + x 2 + · · · + x 10 = 6,subject to x 1 ≥ 1 and x i ≥ 0 when i > 1. Let y 1 = x 1 − 1. Then the above equation isequivalent toy 1 + x 2 + · · · + x 10 = 5,where y 1 ≥ 0 and x i ≥ 0 when i > 1. By the same reasoning as in part (a), there are( ) 14= 14!6 6!8! = 3003possibilities.11. The binomial theorem states thatSetting x = −2 and y = 1, we obtainn∑k=0which is the statement to be proved.(x + y) n =n∑k=0( nk)x k y n−k( nk)(−2) k 1 n−k = (−2 + 1) n = (−1) n ,12. The sample space for this experiment is S = {bbbb, bbbg, bbgb, . . . , gggg}, consisting of 16equally likely outcomes.4

(a) The event A “all boys” contains the single outcome bbbb, and so has probability 1/16.(b) The event “exactly two girls” contains the ( 42)outcomes {ggbb, gbgb, . . . , bbgg}, and sohas probability ( 42)/16 = 6/16 = 3/8.(c) The event B “there is at least one boy” contains all of the possible outcomes, exceptgggg, and so has probability 15/16. We haveP [A|B] =P [A ∩ B]P [B]= P [A]P [B] = 1/1615/16 = 115 .(d) The event “two of one sex, two of the other,” is the same as “exactly two girls,” whichhas probability 3/8.The event “three of one sex, one of the other” contains the 8 outcomes{bbbg, bbgb, bgbb, gbbb, bggg, gbgg, ggbg, gggb},and so has probability 8/16 = 4/8 = 1/2.Clearly the latter event has the greater probability.13. The graph must be finite! In a connected graph of n vertices, each vertex has degree in theset {1, 2, . . . , n − 1}. Let the n vertices be “pigeons” and let the n − 1 possible vertex degreesbe “pigeonholes.” Assign a vertex to a pigeonhole according to its degree. By the pigeonholeprinciple, at least one pigeonhole will contain at least two pigeons, i.e., two vertices will havethe same degree.14. Number the positions in line from 1 to 35. The positions of the 20 hat-wearing studentsforms the sequence h 1 , h 2 , . . . , h 20 of distinct integers, each an element of {1, 2, . . . , 35}. Thesequence h 1 − 4, h 2 − 4, . . . , h 20 − 4 consists of distinct numbers in the range {−3, −2, . . . , 31}.Let the 40 numbers in the two sequences be “pigeons” and let the 39 numbers in the set{−3, −2, . . . , 35} be pigeonholes, where a pigeon is assigned to a pigeonhole according to itsnumber. By the pigeonhole principle, at least two of the 40 numbers must be the same.However, since the numbers h 1 , . . . , h 20 are all distinct, as are the numbers h 1 −4, . . . , h 20 −4,it follows that for some i and j, we have h i = h j − 4. Thus hat-wearing student j is 4-backfrom hat-wearing student i.15. (a) Number the jurors from 1 to 12, and let b i ∈ {1, 2, . . . , 12} denote the birth month ofjuror i, where 1 means January, 2 means February, and so on. There are 12 choices forb 1 , 12 choices for b 2 , etc. By the product rule, there are 12 × 12 × · · · × 12 = 12 12 distinctordered 12-tuples.i. The event “all jurors have birthdays in the same month” consists of the 12 outcomes{(1, 1, . . . , 1), (2, 2, . . . , 2), . . . , (12, 12, . . . , 12)}, and so has probability1212 12 = 112 11 .ii. The event “the jurors have birthdays all in different months” consists of the 12!outcomes that are all the possible permutations of (1, 2, 3, . . . , 12). This event hasprobability12!12 12 .5

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